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How do you differentiate #y = e^(2x) ln x#?

1 Answer

Shwetank Mauria

Sep 28, 2016

#(dy)/(dx)=e^(2x)(2lnx+1/x)#

Explanation:

To differentiate #y=e^(2x)lnx#, we use product rule according to which if #f(x)=g(x)xxh(x)#

#(df)/(dx)=g(x)xx(dh)/(dx)+(dg)/(dx)xxh(x)#

As such #(dy)/(dx)=(e^(2x)xx2xxlnx)+(e^(2x)xx1/x)#

= #e^(2x)(2lnx+1/x)#

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